[Leetcode] 133. Clone Graph 解题报告

最新推荐文章于 2021-05-14 10:42:19 发布

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#Leetcode #解题报告 #BFS #DFS

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本文介绍了一种使用深度优先搜索(DFS)克隆无向图的方法。通过建立哈希表存储已克隆节点，避免了图中可能出现的回路问题。文章提供了简洁易懂的C++代码实现。

题目：

Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.

OJ's undirected graph serialization:

Nodes are labeled uniquely.

We use # as a separator for each node, and , as a separator for node label and each neighbor of the node.

As an example, consider the serialized graph {0,1,2#1,2#2,2}.

The graph has a total of three nodes, and therefore contains three parts as separated by #.

First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
Second node is labeled as 1. Connect node 1 to node 2.
Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.

Visually, the graph looks like the following:

思路：

解决图的问题的最大神器依然是DFS和BFS。由于图有可能会形成回路，所以我们建立一个哈希表来存储已经clone后的节点。在本题目中，我们采用DFS的思路，代码十分简练和易懂。

代码：

/**
 * Definition for undirected graph.
 * struct UndirectedGraphNode {
 *     int label;
 *     vector<UndirectedGraphNode *> neighbors;
 *     UndirectedGraphNode(int x) : label(x) {};
 * };
 */
class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if(node == NULL) {
            return NULL;
        }
        if(hash.count(node)) {
            return hash[node];
        }
        hash[node] = new UndirectedGraphNode(node->label);
        for(auto val : node->neighbors) {
            hash[node]->neighbors.push_back(cloneGraph(val));
        }
        return hash[node];
    }
private:
    unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> hash;
};